7.2: Panua na Gawanya Maneno ya busara
- Page ID
- 176752
Mwishoni mwa sehemu hii, utaweza:
- Kuamua maadili ambayo kujieleza kwa busara haijulikani
- Kurahisisha maneno ya busara
- Kuzidisha maneno ya busara
- Gawanya maneno ya busara
- Panua na ugawanye kazi za busara
Tulipitia upya mali ya sehemu ndogo na shughuli zao. Tulianzisha namba za busara, ambazo ni sehemu ndogo tu ambapo nambari na denominators ni integers. Katika sura hii, tutafanya kazi na sehemu ambazo nambari na denominators ni polynomials. Tunaita aina hii ya kujieleza kuwa kujieleza kwa busara.
Maneno ya busara ni usemi wa fomu\(\dfrac{p}{q}\), where \(p\) na\(q\) ni polynomials na\(q\neq 0\).
Hapa ni baadhi ya mifano ya maneno ya busara:
\[−\dfrac{24}{56} \qquad \dfrac{5x}{12y} \qquad \dfrac{4x+1}{x^2−9} \qquad \dfrac{4x^2+3x−1}{2x−8}\nonumber\]
Kumbuka kwamba kwanza busara kujieleza waliotajwa hapo juu\(−\dfrac{24}{56}\),, ni sehemu tu. Kwa kuwa mara kwa mara ni polynomial na sifuri shahada, uwiano wa constants mbili ni kujieleza busara, mradi denominator si sifuri.
Tutafanya shughuli sawa na maneno ya busara ambayo tulifanya kwa sehemu ndogo. Sisi kurahisisha, kuongeza, Ondoa, kuzidisha, kugawanya na kuitumia katika programu.
Tambua Maadili ambayo Ufafanuzi wa busara haukufafanuliwa
Ikiwa denominator ni sifuri, kujieleza kwa busara haijulikani. Nambari ya kujieleza kwa busara inaweza kuwa 0—lakini sio denominator.
Tunapofanya kazi na sehemu ya namba, ni rahisi kuepuka kugawa kwa sifuri kwa sababu tunaweza kuona namba katika denominator. Ili kuepuka kugawa kwa sifuri kwa kujieleza kwa busara, hatupaswi kuruhusu maadili ya kutofautiana ambayo itafanya denominator kuwa sifuri.
Hivyo kabla ya kuanza operesheni yoyote na kujieleza mantiki, sisi kuchunguza ni kwanza kupata maadili ambayo kufanya denominator sifuri. Kwa njia hiyo, tunapotatua usawa wa busara kwa mfano, tutajua kama ufumbuzi wa algebraic tunayopata unaruhusiwa au la.
- Weka denominator sawa na sifuri.
- Kutatua equation.
Tambua thamani ambayo kila kujieleza kwa busara haijulikani:
a.\(\dfrac{8a^2b}{3c}\) b.\(\dfrac{4b−3}{2b+5}\) c\(\dfrac{x+4}{x^2+5x+6}\).
Suluhisho
Maneno hayatakuwa na ufafanuzi wakati denominator ni sifuri.
a.
\(\begin{array} {ll} &\dfrac{8a^2b}{3c} \\ \begin{array} {l} \text{Set the denominator equal to zero and solve} \\ \text{for the variable.} \end{array} &3c=0 \\ &c=0 \\ &\dfrac{8a^2b}{3c}\text{ is undefined for }c=0 \end{array} \)
b.
\(\begin{array} {ll} &\dfrac{4b-3}{2b+5} \\ \begin{array} {l} \text{Set the denominator equal to zero and solve} \\ \text{for the variable.} \end{array} &2b+5=0 \\ &2b=-5 \\ &b=-\dfrac{5}{2} \\ & \\ &\dfrac{4b-3}{2b+5} \text{ is undefined for }b=-\dfrac{5}{2} \end{array} \)
c.
\(\begin{array} {ll} &\dfrac{x+4}{x^2 + 5x + 6} \\ \begin{array} {l} \text{Set the denominator equal to zero and solve } \\ \text{for the variable.} \end{array} &x^2+5x+6=0 \\ &(x+2)(x+3)=0 \\ &x+2=0\text{ or }x+3=0 \\ &x=-2\text{ or }x=-3 \\ & \\ &\dfrac{x+4}{x^2+5x+6}\text{ is undefined for }x=-2\text{ or }x=-3 \end{array} \)
Tambua thamani ambayo kila kujieleza kwa busara haijulikani.
a.\(\dfrac{3y^2}{8x}\) b.\(\dfrac{8n−5}{3n+1}\) c.\(\dfrac{a+10}{a^2+4a+3}\)
- Jibu
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a.\(x=0\)
b.\(n=−\dfrac{1}{3}\)
c.\(a=−1,a=−3\)
Tambua thamani ambayo kila kujieleza kwa busara haijulikani.
a.\(\dfrac{4p}{5q}\) b.\(\dfrac{y−1}{3y+2}\) c.\(\dfrac{m−5}{m^2+m−6}\)
- Jibu
-
a.\(q=0\)
b.\(y=−\dfrac{2}{3}\)
c.\(m=2,m=−3\)
Kurahisisha maneno ya busara
Sehemu inachukuliwa kuwa rahisi ikiwa hakuna mambo ya kawaida, isipokuwa 1, katika nambari yake na denominator. Vile vile, kujieleza rahisi kwa busara hauna mambo ya kawaida, isipokuwa 1, katika nambari yake na denominator.
Maneno ya busara yanachukuliwa kuwa rahisi ikiwa hakuna mambo ya kawaida katika nambari yake na denominator.
Kwa mfano,
\[ \begin{array} {l} \dfrac{x+2}{x+3} \text{ is simplified because there are no common factors of } x+2 \text{ and }x+3. \\ \dfrac{2x}{3x} \text{ is not simplified because x is a common factor of }2x\text{ and }3x. \\ \end{array} \nonumber\]
Tunatumia Mali sawa FRACTIONS ili kurahisisha sehemu ndogo za namba. Tunarudia tena hapa kama tutakayotumia pia ili kurahisisha maneno ya busara.
Kama\(a\),\(b\), na\(c\) ni idadi ambapo\(b\neq 0,c\neq 0,\)
\[\text {then } \dfrac{a}{b}=\dfrac{a·c}{b·c} \text{ and } \dfrac{a·c}{b·c}=\dfrac{a}{b}\nonumber\]
Ona kwamba katika Mali sawa Fractions, maadili ambayo ingeweza kufanya denominators sifuri ni hasa haruhusiwi. Tunaona\(b\neq 0,c\neq 0\) wazi.
Ili kurahisisha maneno ya busara, sisi kwanza tunaandika nambari na denominator katika fomu iliyopangwa. Kisha sisi kuondoa mambo ya kawaida kwa kutumia sawa Fractions Mali.
Kuwa makini sana kama wewe kuondoa mambo ya kawaida. Mambo yanaongezeka ili kufanya bidhaa. Unaweza kuondoa sababu kutoka kwa bidhaa. Huwezi kuondoa muda kutoka kwa jumla.
Kuondoa ya\(x\) kutoka\(\dfrac{x+5}{x}\) itakuwa kama kufuta katika sehemu\(2\)\(\dfrac{2+5}{2}!\)
Jinsi ya kurahisisha kujieleza kwa busara
Kurahisisha:\(\dfrac{x^2+5x+6}{x^2+8x+12}\)
Suluhisho
Kurahisisha:\(\dfrac{x^2−x−2}{x^2−3x+2}\).
- Jibu
-
\(\dfrac{x+1}{x−1},x\neq 2,x\neq 1\)
Kurahisisha:\(\dfrac{x^2−3x−10}{x^2+x−2}\).
- Jibu
-
\(\dfrac{x−5}{x−1},x\neq −2,x\neq 1\)
Sasa tunafupisha hatua unazopaswa kufuata ili kurahisisha maneno ya busara.
- Fanya namba na denominator kabisa.
- Kurahisisha kwa kugawa mambo ya kawaida.
Kawaida, tunaacha maneno rahisi ya busara katika fomu iliyopangwa. Kwa njia hii, ni rahisi kuangalia kwamba tumeondoa mambo yote ya kawaida.
Tutatumia mbinu tumejifunza kwa sababu ya polynomials katika numerators na denominators katika mifano ifuatayo.
Kila wakati sisi kuandika kujieleza busara, tunapaswa kutoa taarifa disallowing maadili ambayo kufanya denominator sifuri. Hata hivyo, ili tuangalie kazi iliyopo, tutaacha kuandika katika mifano.
Kurahisisha:\(\dfrac{3a^2−12ab+12b^2}{6a^2−24b^2}\).
Suluhisho
\(\begin{array} {ll} &\dfrac{3a^2−12ab+12b^2}{6a^2−24b^2} \\ & \\ & \\ \begin{array} {l} \text{Factor the numerator and denominator,} \\ \text{first factoring out the GCF.} \end{array} &\dfrac{3(a^2−4ab+4b^2)}{6(a^2−4b^2)} \\ & \\ &\dfrac{3(a−2b)(a−2b)}{6(a+2b)(a−2b)} \\ & \\ \text{Remove the common factors of }a−2b\text{ and }3. &\dfrac{\cancel{3}(a−2b)\cancel{(a−2b)}}{\cancel{3}·2(a+2b)\cancel{(a−2b)}} \\ &\dfrac{a−2b}{2(a+2b)} \end{array} \)
Kurahisisha:\(\dfrac{2x^2−12xy+18y^2}{3x^2−27y^2}\).
- Jibu
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\(\dfrac{2(x−3y)}{3(x+3y)}\)
Kurahisisha:\(\dfrac{5x^2−30xy+25y^2}{2x^2−50y^2}\).
- Jibu
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\(\dfrac{5(x−y)}{2(x+5y)}\)
Sasa tutaona jinsi ya kurahisisha kujieleza kwa busara ambao nambari na denominator zina mambo tofauti. Sisi hapo awali ilianzisha notation kinyume: kinyume cha\(a\) ni\(−a\) na\(−a=−1·a\).
sehemu ya namba, kusema\(\dfrac{7}{−7}\) simplifies kwa\(−1\). Pia tunatambua kwamba nambari na denominator ni kinyume.
Sehemu\(\dfrac{a}{−a}\), ambayo nambari yake na denominator ni kinyume pia inaeleza\(−1\).
\[\begin{array} {ll} \text{Let’s look at the expression }b−a. &b−a \\ \text{Rewrite.} &−a+b \\ \text{Factor out }–1. &−1(a−b) \nonumber\end{array} \]
Hii inatuambia kwamba\(b−a\) ni kinyume cha\(a−b\).
Kwa ujumla, tunaweza kuandika kinyume cha\(a−b\) kama\(b−a\). Hivyo kujieleza kwa busara\(\dfrac{a−b}{b−a}\) kurahisisha\(−1\).
Kinyume cha\(a−b\) ni\(b−a\).
\[\dfrac{a−b}{b−a}=−1 \quad a\neq b\nonumber\]
Maneno na mgawanyiko wake kinyume na\(−1\).
Tutatumia mali hii ili kurahisisha maneno ya busara ambayo yana kinyume katika nambari zao na denominators. Kuwa makini si kutibu\(a+b\) na\(b+a\) kama kupinga. Kumbuka kwamba kwa kuongeza, ili haijalishi hivyo\(a+b=b+a\). Hivyo kama\(a\neq −b\), basi\(\dfrac{a+b}{b+a}=1\).
Kurahisisha:\(\dfrac{x^2−4x−32}{64−x^2}\)
Suluhisho
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| Fanya namba na denominator. |  |
| Tambua mambo ambayo yanapinga. |  |
| Kurahisisha. |  |
Kurahisisha:\(\dfrac{x^2−4x−5}{25−x^2}\)
- Jibu
-
\(−\dfrac{x+1}{x+5}\)
Kurahisisha:\(\dfrac{x^2+x−2}{1−x^2}\).
- Jibu
-
\(−\dfrac{x+2}{x+1}\)
Kuzidisha maneno ya busara
Ili kuzidisha maneno ya busara, tunafanya tu kile tulichofanya na sehemu ndogo za namba. Sisi kuzidisha nambari na kuzidisha denominators. Kisha, ikiwa kuna mambo yoyote ya kawaida, tunawaondoa ili kurahisisha matokeo.
Ikiwa\(p\),\(q\),\(r\), na\(s\) ni polynomials ambapo\(q\neq 0\)\(s\neq 0\), basi
\[\dfrac{p}{q}·\dfrac{r}{s}=\dfrac{pr}{qs}\nonumber\]
Ili kuzidisha maneno ya busara, kuzidisha nambari na kuzidisha denominators.
Kumbuka, katika sura hii, tutafikiri kwamba maadili yote ya namba ambayo yangeweza kufanya denominator kuwa sifuri hutolewa. Hatuwezi kuandika vikwazo kwa kila kujieleza kwa busara, lakini kukumbuka kwamba denominator haiwezi kamwe kuwa sifuri. Hivyo katika mfano huu ijayo,\(x\neq 0\),\(x\neq 3\), na\(x\neq 4.\)
Kurahisisha:\(\dfrac{2x}{x^2−7x+12}·\dfrac{x^2−9}{6x^2}\).
Suluhisho
Kurahisisha:\(\dfrac{5x}{x^2+5x+6}·\dfrac{x^2−4}{10x}\).
- Jibu
-
\(\dfrac{x−2}{2(x+3)}\)
Kurahisisha:\(\dfrac{9x^2}{x^2+11x+30}·\dfrac{x^2−36}{3x^2}\).
- Jibu
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\(\dfrac{3(x−6)}{x+5}\)
- Factor kila nambari na denominator kabisa.
- Kuzidisha nambari na denominators.
- Kurahisisha kwa kugawa mambo ya kawaida.
Kuzidisha:\(\dfrac{3a^2−8a−3}{a^2−25}·\dfrac{a^2+10a+25}{3a^2−14a−5}\).
Suluhisho
\(\begin{array} {ll} &\dfrac{3a^2−8a−3}{a^2−25}·\dfrac{a^2+10a+25}{3a^2−14a−5} \\ & \\ \begin{array} {ll} \text{Factor the numerators and denominators} \\ \text{and then multiply.} \end{array} &\dfrac{(3a+1)(a−3)(a+5)(a+5)}{(a−5)(a+5)(3a+1)(a−5)} \\ & \\ \begin{array} {l} \text{Simplify by dividing out} \\ \text{common factors.} \end{array} &\dfrac{\cancel{(3a+1)}(a−3)\cancel{(a+5)}(a+5)}{(a−5)\cancel{(a+5)}\cancel{(3a+1)}(a−5)} \\ & \\ \text{Simplify.} &\dfrac{(a−3)(a+5)}{(a−5)(a−5)} \\ & \\ \text{Rewrite }(a−5)(a−5)\text{ using an exponent.} &\dfrac{(a−3)(a+5)}{(a−5)^2} \end{array}\)
Kurahisisha:\(\dfrac{2x^2+5x−12}{x^2−16}·\dfrac{x^2−8x+16}{2x^2−13x+15}\).
- Jibu
-
\(\dfrac{x−4}{x−5}\)
Kurahisisha:\(\dfrac{4b^2+7b−2}{1−b^2}·\dfrac{b^2−2b+1}{4b^2+15b−4}\).
- Jibu
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\(−\dfrac{(b+2)(b−1)}{(1+b)(b+4)}\)
Gawanya Maneno ya busara
Kama tulivyofanya kwa vipande vya namba, kugawanya maneno ya busara, tunazidisha sehemu ya kwanza kwa usawa wa pili.
Ikiwa\(p\),\(q\),\(r\), na\(s\) ni polynomials ambapo\(q\neq 0\),\(r\neq 0\),\(s\neq 0\), basi
\[\dfrac{p}{q}÷\dfrac{r}{s}=\dfrac{p}{q}·\dfrac{s}{r}\nonumber\]
Ili kugawanya maneno ya busara, kuzidisha sehemu ya kwanza kwa usawa wa pili.
Mara tu tunapoandika tena mgawanyiko kama kuzidisha kwa kujieleza kwanza kwa usawa wa pili, sisi kisha tunafanya kila kitu na kuangalia mambo ya kawaida.
Gawanya:\(\dfrac{p^3+q^3}{2p^2+2pq+2q^2}÷\dfrac{p^2−q^2}{6}\).
Suluhisho
Kurahisisha:\(\dfrac{x^3−8}{3x^2−6x+12}÷\dfrac{x^2-4}{6}\).
- Jibu
-
\(\dfrac{2(x^2+2x+4)}{(x+2)(x^2−2x+4)}\)
Kurahisisha:\(\dfrac{2z^2}{z^2−1}÷\dfrac{z^3−z^2+z}{z^3+1}\).
- Jibu
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\(\dfrac{2z}{z−1}\)
- Andika upya mgawanyiko kama bidhaa ya kujieleza kwa busara ya kwanza na usawa wa pili.
- Factor numerators na denominators kabisa.
- Kuzidisha nambari na denominators pamoja.
- Kurahisisha kwa kugawa mambo ya kawaida.
Kumbuka kutoka Matumizi Lugha ya Algebra kwamba sehemu tata ni sehemu ambayo ina sehemu katika nambari, denominator au wote wawili. Pia, kumbuka bar sehemu ina maana mgawanyiko. Sehemu ngumu ni njia nyingine ya kuandika mgawanyiko wa sehemu mbili.
Gawanya:\(\dfrac{\dfrac{6x^2−7x+2}{4x−8}}{\dfrac{2x^2−7x+3}{x^2−5x+6}}\).
Suluhisho
\(\begin{array} {ll} &\dfrac{\dfrac{6x^2−7x+2}{4x−8}}{\dfrac{2x^2−7x+3}{x^2−5x+6}} \\ & \\ \text{Rewrite with a division sign.} &\dfrac{6x^2−7x+2}{4x−8}÷\dfrac{2x^2−7x+3}{x^2−5x+6} \\ & \\ \begin{array} {l} \text{Rewrite as product of first times reciprocal} \\ \text{of second.} \end{array} &\dfrac{6x^2−7x+2}{4x−8}·\dfrac{x^2−5x+6}{2x^2−7x+3} \\ & \\ \begin{array} {l} \text{Factor the numerators and the} \\ \text{denominators, and then multiply.} \end{array} &\dfrac{(2x−1)(3x−2)(x−2)(x−3)}{4(x−2)(2x−1)(x−3)} \\ & \\ \text{Simplify by dividing out common factors.} &\dfrac{\cancel{(2x−1)}(3x−2)\cancel{(x−2)}\cancel{(x−3)}}{4\cancel{(x−2)}\cancel{(2x−1)}\cancel{(x−3)}} \\ \text{Simplify.} &\dfrac{3x−2}{4} \end{array}\)
Kurahisisha:\(\dfrac{\dfrac{3x^2+7x+2}{4x+24}}{\dfrac{3x^2−14x−5}{x^2+x−30}}\).
- Jibu
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\(\dfrac{x+2}{4}\)
Kurahisisha:\(\dfrac{\dfrac{y^2−36}{2y^2+11y−6}}{\dfrac{2y^2−2y−60}{8y−4}}\).
- Jibu
-
\(\dfrac{2}{y+5}\)
Ikiwa tuna maneno zaidi ya mbili ya busara ya kufanya kazi na, bado tunafuata utaratibu huo. Hatua ya kwanza itakuwa kuandika tena mgawanyiko wowote kama kuzidisha kwa usawa. Kisha, tunazingatia na kuzidisha.
Fanya shughuli zilizoonyeshwa:\(\dfrac{3x−6}{4x−4}·\dfrac{x^2+2x−3}{x^2−3x−10}÷\dfrac{2x+12}{8x+16}\).
Suluhisho
 |
|
| Andika upya mgawanyiko kama kuzidisha kwa usawa. |
 |
| Factor numerators na denominators. |  |
| Panua sehemu ndogo. Kuleta mara kwa mara mbele itasaidia wakati wa kuondoa mambo ya kawaida. |
|
| Kurahisisha kwa kugawa mambo ya kawaida. |  |
| Kurahisisha. |  |
Fanya shughuli zilizoonyeshwa:\(\dfrac{4m+4}{3m−15}·\dfrac{m^2−3m−10}{m^2−4m−32}÷\dfrac{12m−36}{6m−48}\).
- Jibu
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\(\dfrac{2(m+1)(m+2)}{3(m+4)(m−3)}\)
Fanya shughuli zilizoonyeshwa:\(\dfrac{2n^2+10n}{n−1}÷\dfrac{n^2+10n+24}{n^2+8n−9}·\dfrac{n+4}{8n^2+12n}\).
- Jibu
-
\(\dfrac{(n+5)(n+9)}{2(n+6)(2n+3)}\)
Kuzidisha na Kugawanya Kazi
Tulianza sehemu hii na kusema kuwa kujieleza busara ni usemi wa fomu\(\dfrac{p}{q}\), ambapo p na q ni polynomials na\(q\neq 0\). Vile vile, tunafafanua kazi ya busara kama kazi ya fomu\(R(x)=\dfrac{p(x)}{q(x)}\) ambapo\(p(x)\) na\(q(x)\) ni kazi nyingi na\(q(x)\) sio sifuri.
Kazi ya busara ni kazi ya fomu
\[R(x)=\dfrac{p(x)}{q(x)}\nonumber\]
wapi\(p(x)\) na\(q(x)\) ni kazi polynomial na\(q(x)\) si sifuri.
Kikoa cha kazi ya busara ni namba zote halisi isipokuwa kwa maadili hayo ambayo yangeweza kusababisha mgawanyiko kwa sifuri. Lazima tuondoe maadili yoyote ambayo hufanya\(q(x)=0\).
- Weka denominator sawa na sifuri.
- Kutatua equation.
- Kikoa ni namba zote halisi ukiondoa maadili yaliyopatikana katika Hatua ya 2.
Kupata uwanja wa\(R(x)=\dfrac{2x^2−14x}{4x^2−16x−48}\).
Suluhisho
Kikoa kitakuwa namba zote halisi isipokuwa maadili hayo yanayofanya denominator sifuri. Tutaweka denominator sawa na sifuri, kutatua equation hiyo, na kisha uondoe maadili hayo kutoka kwenye kikoa.
\(\begin{array} {ll} \text{Set the denominator to zero.} &4x^2−16x−48=0 \\ \text{Factor, first factor out the GCF.} &4(x^2−4x−12)=0 \\ &4(x−6)(x+2)=0 \\ \text{Use the Zero Product Property.} &4\neq 0\quad x−6=0\quad x+2=0 \\ \text{Solve.} &\hspace{24mm}x=6\qquad x=−2 \\ &\text{The domain of }R(x)\text{ is all real numbers} \\ &\text{where }x\neq 6\text{ and }x\neq −2 \end{array}\).
Kupata uwanja wa\(R(x)=\dfrac{2x^2−10x}{4x^2−16x−20}\).
- Jibu
-
uwanja wa\(R(x)\) ni namba zote halisi ambapo\(x\neq 5\) na\(x\neq −1\).
Kupata uwanja wa\(R(x)=\dfrac{4x^2−16x}{8x^2−16x−64}\).
- Jibu
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uwanja wa\(R(x)\) ni namba zote halisi ambapo\(x\neq 4\) na\(x\neq −2\).
Ili kuzidisha kazi za busara, tunazidisha maneno ya busara yanayosababisha upande wa kulia wa equation kwa kutumia mbinu sawa ambazo tulizitumia kuzidisha maneno ya busara.
Pata\(R(x)=f(x)·g(x)\) wapi\(f(x)=\dfrac{2x−6}{x^2−8x+15}\) na\(g(x)=\dfrac{x^2−25}{2x+10}\).
Suluhisho
\(\begin{array} {ll} &R(x)=f(x)·g(x) \\ & \\ &R(x)=\dfrac{2x−6}{x^2−8x+15}·\dfrac{x^2−25}{2x+10} \\ & \\ \text{Factor each numerator and denominator.} &R(x)=\dfrac{2(x−3)}{(x−3)(x−5)}·\dfrac{(x−5)(x+5)}{2(x+5)} \\ & \\ \text{Multiply the numerators and denominators.} &R(x)=\dfrac{2(x−3)(x−5)(x+5)}{2(x−3)(x−5)(x+5)} \\ & \\ \text{Remove common factors.} &R(x)=\dfrac{\cancel{2}\cancel{(x−3)}\cancel{(x−5)}\cancel{(x+5)}}{\cancel{2}\cancel{(x−3)}\cancel{(x−5)}\cancel{(x+5)}} \\ & \\ \text{Simplify.} &R(x)=1 \end{array}\)
Pata\(R(x)=f(x)·g(x)\) wapi\(f(x)=\dfrac{3x−21}{x^2−9x+14}\) na\(g(x)=\dfrac{2x^2−8}{3x+6}\).
- Jibu
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\(R(x)=2\)
Pata\(R(x)=f(x)·g(x)\) wapi\(f(x)=\dfrac{x^2−x}{3x^2+27x−30}\) na\(g(x)=\dfrac{x^2−100}{x^2−10x}\).
- Jibu
-
\(R(x)=\dfrac{1}{3}\)
Ili kugawanya kazi za busara, tunagawanya maneno ya busara yanayosababisha upande wa kulia wa equation kwa kutumia mbinu sawa ambazo tulizitumia kugawanya maneno ya busara.
Pata\(R(x)=\dfrac{f(x)}{g(x)}\) wapi\(f(x)=\dfrac{3x^2}{x^2−4x}\) na\(g(x)=\dfrac{9x^2−45x}{x^2−7x+10}\).
Suluhisho
\(\begin{array} {ll} &R(x)=\dfrac{f(x)}{g(x)} \\ \text{Substitute in the functions }f(x),\space g(x). &R(x)=\dfrac{\dfrac{3x^2}{x^2−4x}}{\dfrac{9x^2−45x}{x^2−7x+10}} \\ & \\ \begin{array} {l} \text{Rewrite the division as the product of} \\ f(x)\text{ and the reciprocal of }g(x). \end{array} &R(x)=\dfrac{3x^2}{x^2−4x}·\dfrac{x^2−7x+10}{9x^2−45x} \\ & \\ \begin{array} {l} \text{Factor the numerators and denominators} \\ \text{and then multiply.} \end{array} &R(x)=\dfrac{3·x·x·(x−5)(x−2)}{x(x−4)·3·3·x·(x−5)} \\ & \\ \text{Simplify by dividing out common factors.} &R(x)=\dfrac{\cancel{3}·\cancel{x}·\cancel{x}\cancel{(x−5)}(x−2)}{\cancel{x}(x−4)·\cancel{3}·3·\cancel{x}\cancel{(x−5)}} \\ & \\ &R(x)=\dfrac{x−2}{3(x−4)} \end{array}\)
Pata\(R(x)=\dfrac{f(x)}{g(x)}\) wapi\(f(x)=\dfrac{2x^2}{x^2−8x}\) na\(g(x)=\dfrac{8x^2+24x}{x^2+x−6}\).
- Jibu
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\(R(x)=\dfrac{x−2}{4(x−8)}\)
Pata\(R(x)=\dfrac{f(x)}{g(x)}\) wapi\(f(x)=\dfrac{15x^2}{3x^2+33x}\) na\(g(x)=\dfrac{5x−5}{x^2+9x−22}\).
- Jibu
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\(R(x)=\dfrac{x(x−2)}{x−1}\)
Dhana muhimu
- Kuamua maadili ambayo kujieleza kwa busara haijulikani.
- Weka denominator sawa na sifuri.
- Kutatua equation.
- Sawa Fractions Mali
Kama\(a\)\(b\),, na\(c\) ni idadi ambapo\(b\neq 0\),\(c\neq 0\), basi
\(\quad\dfrac{a}{b}=\dfrac{a·c}{b·c}\) na\(\dfrac{a·c}{b·c}=\dfrac{a}{b}.\) - Jinsi ya kurahisisha kujieleza kwa busara.
- Fanya namba na denominator kabisa.
- Kurahisisha kwa kugawa mambo ya kawaida.
- Vikwazo katika kujieleza
Mantiki kinyume cha\(a−b\) ni\(b−a\).
\(\quad\dfrac{a−b}{b−a}=−1 \qquad a\neq b\)
Maneno na mgawanyiko wake kinyume na\(−1\). - Kuongezeka kwa Maneno ya busara
Ikiwa\(p\)\(q\),\(r\),, na\(s\) ni polynomials ambapo\(q\neq 0\)\(s\neq 0\), basi
\(\quad\dfrac{p}{q}·\dfrac{r}{s}=\dfrac{pr}{qs}\) - Jinsi ya kuzidisha maneno ya busara.
- Factor kila nambari na denominator kabisa.
- Kuzidisha nambari na denominators.
- Kurahisisha kwa kugawa mambo ya kawaida.
- Idara ya Maneno ya busara
Ikiwa\(p\)\(q\),\(r\),, na\(s\) ni polynomials ambapo\(q\neq 0\)\(r\neq 0\),,\(s\neq 0\), basi
\(\quad\dfrac{p}{q}÷\dfrac{r}{s}=\dfrac{p}{q}·\dfrac{s}{r}\) - Jinsi ya kugawanya maneno ya busara.
- Andika upya mgawanyiko kama bidhaa ya kujieleza kwa busara ya kwanza na usawa wa pili.
- Factor numerators na denominators kabisa.
- Kuzidisha nambari na denominators pamoja.
- Kurahisisha kwa kugawa mambo ya kawaida.
- Jinsi ya kuamua uwanja wa kazi ya busara.
- Weka denominator sawa na sifuri.
- Kutatua equation.
- Kikoa ni namba zote halisi ukiondoa maadili yaliyopatikana katika Hatua ya 2.
faharasa
- kujieleza kwa busara
- Maneno ya busara ni usemi wa fomu\(\dfrac{p}{q}\), wapi\(p\) na\(q\) ni polynomials na\(q\neq 0\).
- rahisi kujieleza busara
- Maneno rahisi ya busara hayana mambo ya kawaida, isipokuwa\(1\), katika nambari yake na denominator.
- kazi ya busara
- Kazi ya busara ni kazi ya fomu\(R(x)=\dfrac{p(x)}{q(x)}\) ambapo\(p(x)\) na\(q(x)\) ni kazi nyingi na\(q(x)\) sio sifuri.